@AlecS Better, but the substitution $2rN=L$ seems to be based on the assumption that the windings are all in a single layer. It is quite common to wind the wire in several layers. Thereby, increasing the number of turns does not necessarily increase $L$. It would seem more obvious to make the substitution $l=Ns$, where $s$ is the length of each turn, since increasing the number of turns will most definitely make the wire longer.
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jkejApr 5 '13 at 18:41

When you're increasing the voltage, you're increasing the current. Recall Ohm's law. Provided a steady current flows through the material at constant temperature, the resistance remains the same. If $I$ is increased, then $V$ has to increase in order for the resistance to remain constant.

The magnetic field along the axis of the solenoid is given by $$B=\mu \frac{NI}{L}$$

The permeability of the medium $\mu$, which can be of the order of some $10^5$ when the coil is wound on a soft iron core, which is really helpful. The current (or voltage) can be increased.

There's a problem with increasing the number of turns. When solving for Ohm's law, the expression when you combine $I=nAev_d$ and $v_d=eE\tau/m$, you get
$$V=\frac{mL}{nAe^2\tau}I$$

The large expression in the middle is the resistance $R$. It can be seen that $R\ \alpha\ l/A$. It can be seen that the larger the area, the less resistance there is. So if you're using a long coil, make sure that the area occupied by the coil per turn is more. There's another problem if you're using an AC. The number of turns can also increase the inductance of the coil and hence may lead to impedance, which also adds up with the resistance.

Hence, I suggest the use of soft iron core, increase the current and voltage input, use a wire with low specific resistivity $\rho$ like silver or aluminum. And finally - if you require more number of turns, use thick wires and make sure that the coil occupies a larger area.

The $B$ field does matter, but the force that is exerted from the field is $$F=qVB\ ,$$
where $q$ is charge, $V$ is voltage, and $B$ is magnetic field. Given the equation for $B\ ,$
$$B=\frac{\mu N V }{L R}\ ,$$
we have that $$F= \frac{V^2 N \mu}{L R}\ .$$Thus, in order to optimize the force output you must use voltage more than anything.